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A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red (ii) white (iii) not green |
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Answer» Total number of possible outcomes, n(S) = 5 + 8 + 4 = 17 (i) Number of favorable outcomes, n(E) = 5 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{5}{17}\) (ii) Number of favorable outcomes, n(E) = 8 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{8}{17}\) (iii) Number of events of drawing a green marble = 4 ∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{4}{17}\) Probability of not drawing green marble = 1 – P(E) = 1 - \(\frac{4}{17}\) = \(\frac{13}{17}\) |
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